The answer is \(\sqrt{2}\operatorname{cis} \frac{5\pi}{9}\).
To see this, we first express \(\sqrt{2}-\sqrt{6}i\) in polar form
as \(r \operatorname{cis} \theta\) where \(r\) is the modulus of
\(\sqrt{2} - \sqrt{6}i\) and \(\theta\) satisfies \(0 \leq \theta < 2\pi\).
Now, \(r = \sqrt{ \left(\sqrt{2}\right)^2 +\left(-\sqrt{6}\right)^2 }
= \sqrt{2+6} = \sqrt{8} = \left(\sqrt{2}\right)^3\)
and \(\theta = 2\pi - \tan^{-1} \frac{\sqrt{6}}{\sqrt{2}}
= 2\pi - \tan^{-1} \sqrt{3} = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}\).
Hence,
\(\sqrt{2}-\sqrt{6}i = \left(\sqrt{2}\right)^3\operatorname{cis}\frac{5\pi}{3}\).
We can now proceed in one of two ways:
-
We list all the cube roots of
\(\left(\sqrt{2}\right)^3\operatorname{cis}\frac{5\pi}{3}\):
\(\sqrt{2}\operatorname{cis}\left(\frac{1}{3}\cdot\frac{5\pi}{3}\right)\),
\(\sqrt{2}\operatorname{cis}\left(\frac{1}{3}\cdot\left(2\pi+\frac{5\pi}{3}\right)\right)\),
\(\sqrt{2}\operatorname{cis}\left(\frac{1}{3}\cdot\left(4\pi+\frac{5\pi}{3}\right)\right)\),
which simplify to
\(\sqrt{2}\operatorname{cis}\frac{5\pi}{9}\),
\(\sqrt{2}\operatorname{cis}\frac{11\pi}{9}\),
\(\sqrt{2}\operatorname{cis}\frac{17\pi}{9}\).
Only \(\sqrt{2}\operatorname{cis}\frac{5\pi}{9}\) appears in the list of choices.
-
We use de Moivre's Theorem to compute the cubes of the given choices
and see which one is equal to \(\left(\sqrt{2}\right)^3\operatorname{cis}\frac{5\pi}{3}\).